\(25sin^2x+15sin2x+9cos^2x=25\\\Leftrightarrow (5sinx)^2+15.2sinxcosx+(3cosx)^2=25\\\Leftrightarrow (5sinx+3cosx)^2=25 \\\Leftrightarrow \left[\begin{array}{I}5sinx+3cosx=5\\5sinx+3cosx=-5\end{array}\right.\\TH1:5sinx+3cosx=5\\\Leftrightarrow \dfrac{5}{\sqrt{34}}sinx+\dfrac{3}{\sqrt{34}}cosx=\dfrac{5}{\sqrt{34}}\\\Leftrightarrow arccos\left(\dfrac{5}{\sqrt{34}}\right)sinx+\arcsin\left(\dfrac{3}{\sqrt{34}}\right)cosx=\dfrac{5}{\sqrt{34}}\\\Leftrightarrow sin\left(x+arcsin\dfrac{3}{\sqrt{34}}\right)=\dfrac{5}{\sqrt{34}}\\\Leftrightarrow \left[\begin{array}{I}x+arcsin\dfrac{3}{\sqrt{34}}=arcsin\dfrac{5}{\sqrt{34}}+k2\pi\\x+arcsin\dfrac{3}{\sqrt{34}}=\pi-arcsin\dfrac{5}{\sqrt{34}}+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=arcsin\dfrac{5}{\sqrt{34}}-arcsin\dfrac{3}{\sqrt{34}}+k2\pi\\x=\pi-arcsin\dfrac{5}{\sqrt{34}}-arcsin\dfrac{3}{\sqrt{34}}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\TH2:5sinx+3cosx=-5\\\Leftrightarrow \dfrac{5}{\sqrt{34}}sinx+\dfrac{3}{\sqrt{34}}cosx=-\dfrac{5}{\sqrt{34}}\\\Leftrightarrow arccos\left(\dfrac{5}{\sqrt{34}}\right)sinx+arcsin\left(\dfrac{3}{\sqrt{34}}\right)cosx=-\dfrac{5}{\sqrt{34}}\\\Leftrightarrow sin\left(x+arcsin\dfrac{3}{\sqrt{34}}\right)=-\dfrac{5}{\sqrt{34}}\\\Leftrightarrow \left[\begin{array}{I}x+arcsin\dfrac{3}{\sqrt{34}}=arcsin\left(-\dfrac{5}{\sqrt{34}}\right)+k2\pi\\x+\arcsin\dfrac{3}{\sqrt{34}}=\pi-arcsin\left(-\dfrac{5}{\sqrt{34}}\right)+k2\pi\end{array}\right.\\\Leftrightarrow \left[\begin{array}{I}x=arcsin\left(-\dfrac{5}{\sqrt{34}}\right)-arcsin\dfrac{3}{\sqrt{34}}+k2\pi\\x=\pi-arcsin\left(-\dfrac{5}{\sqrt{34}}\right)-arcsin\dfrac{3}{\sqrt{34}}+k2\pi.\end{array}\right.\quad(k\in\mathbb{Z})\\\text{Vậy phương trình có các họ nghiệm là:}\\x=arcsin\dfrac{5}{\sqrt{34}}-arcsin\dfrac{3}{\sqrt{34}}+k2\pi;\\x=\pi-arcsin\dfrac{5}{\sqrt{34}}-arcsin\dfrac{3}{\sqrt{34}}+k2\pi;\\x=arcsin\left(-\dfrac{5}{\sqrt{34}}\right)-arcsin\dfrac{3}{\sqrt{34}}+k2\pi;\\x=\pi-arcsin\left(-\dfrac{5}{\sqrt{34}}\right)-arcsin\dfrac{3}{\sqrt{34}}+k2\pi.(k\in\mathbb{Z})\)
